Problem 11 - A Graphical Approach to College Algebra

Question

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. State the domain of \(f .\) $$f(x)=\frac{4-3 x}{2 x+1}$$

Step-by-Step Solution

Answer

Vertical asymptote at \(x = -\frac{1}{2}\), horizontal asymptote at \(y = -\frac{3}{2}\), domain is \((-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)\).

Step 1: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator of the function is zero and the numerator is not zero. Set the denominator equal to zero and solve for \(x\): \[2x + 1 = 0\] Solving gives:\[x = -\frac{1}{2}\]This is where the function has a vertical asymptote.

Step 2: Identify Horizontal Asymptotes

For rational functions of the form \(\frac{a_n x^n + \.\.\.}{b_m x^m + \.\.\.}\), if \(n = m\), the horizontal asymptote is \(y = \frac{a_n}{b_m}\). Here, \(n = m = 1\). The leading coefficients are \(-3\) (numerator) and \(2\) (denominator). Therefore, the horizontal asymptote is:\[ y = \frac{-3}{2} \]

Step 3: Determine Oblique Asymptotes

Oblique asymptotes occur if the degree of the numerator is one greater than the degree of the denominator. In this case, the degrees are the same, so there is no oblique asymptote.

Step 4: State the Domain

The function is undefined where the denominator is zero. From Step 1, we found that \(x = -\frac{1}{2}\) causes the denominator to be zero. Hence, the domain of \(f(x)\) is all real numbers except \(x = -\frac{1}{2}\). In interval notation, this is:\((-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)\)

Key Concepts

Vertical Asymptotes

Vertical asymptotes are an important feature of rational functions, representing lines where the graph of a function approaches but never touches or crosses. These lines are found by identifying the values of \(x\) that make the denominator zero, as long as these do not make the numerator zero at the same time. In the given function, \(f(x)=\frac{4-3x}{2x+1}\), the denominator is \(2x + 1\). To find where it is zero, set up the equation \(2x + 1 = 0\). Solving for \(x\), we find \(x = -\frac{1}{2}\). This tells us there is a vertical asymptote at \(x = -\frac{1}{2}\). Points to remember:

• Vertical asymptotes occur where the denominator is zero and the numerator is not zero.
• Solve \(2x + 1 = 0\) to find \(x = -\frac{1}{2}\).
• The graph will approach but never touch the line \(x = -\frac{1}{2}\).

Vertical asymptotes indicate undefined or infinite behavior in a function and help in sketching the graph accurately.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a rational function as \(x\) approaches infinity. For the function \(f(x)=\frac{4-3x}{2x+1}\), we observe the degrees of the polynomial in the numerator (4-3x) and the denominator (2x+1). Both have degrees of 1. When the degrees are equal, the horizontal asymptote can be found by taking the ratio of the leading coefficients.In this case, the leading coefficient of the numerator is -3, and for the denominator, it is 2. Therefore, the horizontal asymptote is given by:\(y = \frac{-3}{2}\).Some key points to remember:

• If degrees are equal, use the ratio of the leading coefficients to find the horizontal asymptote.
• Here, the horizontal asymptote is \(y = -\frac{3}{2}\).
• As \(x\) approaches infinity, the graph of the function will approach this line.

Horizontal asymptotes help in understanding the end behavior of rational functions.

Domain of a Function

The domain of a rational function is the set of all possible input values \(x\) for which the function is defined. For \(f(x)=\frac{4-3x}{2x+1}\), the function is undefined wherever the denominator is zero, since division by zero is undefined in mathematics.From the analysis in the vertical asymptote section, we established that when \(x = -\frac{1}{2}\), the denominator \(2x + 1\) becomes zero. Hence, the domain of the function is all real numbers except \(x = -\frac{1}{2}\). In interval notation, this is expressed as: \((- \infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)\). Key points about the domain:

• The function is undefined where the denominator equals zero.
• In this case, the exception is \(x = -\frac{1}{2}\). Remove this from the set of real numbers.
• The domain helps identify where a function can and cannot exist.

Understanding the domain is crucial for graphing and solving rational functions accurately.